All Questions
17
questions
1
vote
1
answer
72
views
Proving matrix exponential
Can anyone tell me how the following is derived?
where $A$ is a matrix.
-1
votes
1
answer
41
views
Integrate the solution of a the Matrix differential equation
I have:
$\dot{\textbf{x}}=A{\textbf{x}}$
where A is a nxn matrix.
This equation has solution:
$\textbf{x}(t)=e^{\textbf{A}t}\textbf{x}(0)$
A book i'm reading states that:
$\textbf{x}(t_{2})=e^{\...
- 215
0
votes
0
answers
94
views
exponential of matrix in linear differential equation
I am a bit confused about the following from my lecture note:
Suppose we have
\begin{equation}
\begin{aligned}
\frac{dx}{dt}&=-\epsilon y \\
\frac{dy}{dt}&=by+cx
\end{aligned}
\end{...
- 683
1
vote
2
answers
111
views
Stuck on matrix exponential problem [duplicate]
I want to show that the property $e^{tA}e^{tB}=e^{t(A+B)}$ implies that $AB=BA$. Here $A,B$ denotes matrices and $t\in \mathbb{R}$.
Im stuck, have tried to expand both sides with their taylor series ...
- 159
2
votes
1
answer
257
views
Prove that if $P^{-1}AP=diag[\lambda_j]$ then $e^{At}=Pdiag[e^{\lambda_j t}]P^{-1}$
I'm stuck in this exercise.
If $P^{-1}AP=diag[\lambda_j]$ then $e^{At}=Pdiag[e^{\lambda_j t}]P^{-1}$
This is what I've done:
$$P^{-1}AP=diag[\lambda_j]$$
$$\implies AP=Pdiag[\lambda_j]$$
$$\...
user486983
1
vote
1
answer
41
views
Exponential of a non terminating matric
So I understand how to calculate the exponential of matrices that eventually terminate; however, how to approach the cases in which the matrix does not seem to truncate? For example with the matrix $M=...
- 51
1
vote
2
answers
896
views
How to calculate matrix exponential of a $2\times 2$ matrix with repeated e values
Specifically, I am trying to calculate the matrix exponential, $e^{At}$, where A = $\begin{bmatrix}-1 & 1\\-9 & 5\end{bmatrix}$. I calculated the the E values to be 2 with a multiplicity of 2 ...
- 471
1
vote
1
answer
926
views
solve initial value problem using exponential matrix
$x'' = 2 x' +6y +3$
$y' = -x' -2y$
subject the the initial condition
$x(0) = 0; x'(0) = 0; y(0) = 1$
The first part of the question is about finding $e^{At}$ of this matrix $A = \begin{bmatrix}
...
0
votes
1
answer
111
views
Use given identity to computer exponent of 4x4 matrix
I've been given an identity (that I don't know how to prove unfortunately), and been asked to use it to compute exp$(xM)$, where $$ M =
\begin{bmatrix}
1 & 1 & 1 & 1 \\
...
- 31
0
votes
3
answers
121
views
Exponentiation of a $2\times 2$ matrix
We know:
$$\exp(At)=I+ \sum^{\infty}_{n=1}\frac{A^nt^n}{n!}$$
Here $$A= \begin{pmatrix} 0 & 1 \\ -w^2 & 0\end{pmatrix}$$ is a $2\times 2$ matrix,
$I$ is identity matrix.
How to show:
$$\...
- 1,919
2
votes
4
answers
924
views
Exponential of matrix
So, I'm wondering if there is an easy way (as in not calculating the eigenvalues, Jordan canonical form, change of basis matrix, etc.) to calculate this exponential $e^{At}$
with $$A=\begin{pmatrix} 0&...
- 21
0
votes
2
answers
78
views
Let $A$ be a single $p\times p$ Jordan block. Find general solution to $\dfrac{dx}{dt} = Ax$
Let $A$ be a single $p\times p$ Jordan block. Find the general solution to $\,\dfrac{dx}{dt} = Ax$.
What should I approach first? Please help!
- 117
2
votes
3
answers
123
views
If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, then $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$
If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, show that $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$.
I have already been able to show that if $A$ is an arbitrary $2\...
- 1,369
8
votes
4
answers
7k
views
How to compute time ordered Exponential?
So say you have a matrix dependent on a variable t:
$A(t)$
How do you compute $e^{A(t)}$ ?
It seems Sylvester's formula, my standard method of computing matrix exponentials can't be applied here ...
- 17.5k
4
votes
2
answers
4k
views
Show that $e^{t(A+B)} = e^{tA}e^{tB}$ for all $t \in \mathbb{R}$ if, and only if $AB = BA$.
Let A,B real or complex matrixes. Show that $e^{t(A+B)} = e^{tA}e^{tB}$ for all $t \in \mathbb{R}$ if, and only if $AB = BA$.
I demonstrated the reciprocal:
$\Leftarrow )$ The two equations are ...
- 1,819